how to change the value of a variable inside an if loop each time the if loop is triggered ?

Question

i am trying to use codeacademy to learn python. the assignment is to "Write a function called fizz_count that takes a list x as input and returns the count of the string “fizz” in that list."

# Write your function below!
 def fizz_count(input):
     x = [input]
     count = 0
     if x =="fizz":
         count = count + 1
         return count

i think the code above the if loop is fine since the error message ("Your function fails on fizz_count([u'fizz', 0, 0]); it returns None when it should return 1.") only appears when i add that code.

i also tried to make a new variable (new_count) and set that to count + 1 but that gives me the same error message

I would appreciate your assistance very much

input is already a list. You do not need to nest it inside another list literal. — user1726343
– user1726343, Commented Jul 16, 2013 at 1:27
By the sounds of it, you're just after return input.count('fizz') - you don't need to retain a counter around in your fizz_count function. At the moment your function is always going to fall off the end of the function and return None... — Jon Clements
– Jon Clements, Commented Jul 16, 2013 at 1:28
jon clements, that worked :D do you want to post that as an answer ? so i can upvote and set as solution ? — user1476390
– user1476390, Commented Jul 16, 2013 at 1:30
@Jon: Although input.count does the work for you, it would be more pedagogical to implement it oneself. @user1476390: your implementation needs a for loop, at the very least. Oops, now there's lots of answers that give the whole thing away... — Jim Pivarski
– Jim Pivarski, Commented Jul 16, 2013 at 1:32

jh314 · Accepted Answer · 2013-07-16 01:33:17Z

3

The problem is that you have no loop.

# Write your function below!
 def fizz_count(input):
     count = 0
     for x in input: # you need to iterate through the input list
         if x =="fizz":
             count = count + 1
     return count

There is a more concise way by using the .count() function:

def fizz_count(input):
    return input.count("fizz")

answered Jul 16, 2013 at 1:33

jh314

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2 Comments

user1476390 Over a year ago

would x become a list in the for loop ? I am confused because the instructions said to make a list x as input.

jh314 Over a year ago

x takes on the value of each element in the list input. In this case, in the first loop iteration, x becomes the first element in the list. In the second loop iteration, x becomes the second element in the list, and so on.

arshajii · Accepted Answer · 2013-07-16 01:42:07Z

Get rid of x = [input], that just creates another list containing the list input.

i think the code above the if loop is fine

ifs don't loop; you're probably looking for for:

for x in input:  # 'x' will get assigned to each element of 'input'
    ...

Within this loop, you would check if x is equal to "fizz" and increment the count accordingly (as you are doing with your if-statement currently).

Lastly, move your return-statement out of the loop / if-statement. You want that to get executed after the loop, since you always want to traverse the list entirely before returning.

As a side note, you shouldn't use the name input, as that's already assigned to a built-in function.

Putting it all together:

def fizz_count(l):

    count = 0  # set our initial count to 0

    for x in l:  # for each element x of the list l
        if x == "fizz":  # check if x equals "fizz"
            count = count + 1  # if so, increment count

    return count  # return how many "fizz"s we counted

Kubalskiy · Accepted Answer · 2014-02-12 00:45:38Z

0

def fizz_count(x):                       #DEFine func
    count = 0                            #set counter to zero
    for item in x:                       
        if item == "fizz" :               
            count += 1                   #iterate counter +1 for each match
    print count                          #print result
    return count                         #return value 
fizz_count(["fizz","buzz","fizz"])       #call func

answered Feb 12, 2014 at 0:45

Kubalskiy

12 bronze badges

Comments

Don Cruickshank · Accepted Answer · 2014-03-03 14:46:12Z

0

Try this:

# Write your function below!
def fizz_count(x):
    count = 0
    for item in x:
        if item == "fizz": 
            count = count + 1
    return count

edited Mar 3, 2014 at 14:46

Don Cruickshank

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answered Mar 3, 2014 at 14:23

user3375241

1

Comments

matt3141 · Accepted Answer · 2013-07-16 03:48:25Z

-2

def fizz_count(input)
    count = 0
    for x in input:
        count +=  1 if x=="fizz" else 0
    return count

edited Jul 16, 2013 at 3:48

answered Jul 16, 2013 at 1:30

matt3141

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3 Comments

Rushy Panchal Over a year ago

You could make this working by changing count += x=="fizz" ? 1 : 0 to count += (x=="fizz"). If it is True, it is evaluated as 1, else 0. To make it more readable, however, you can change it to: count += 1 if x=="fizz" else 0.

arshajii Over a year ago

I would prefer a simple if-statement to both of those alternatives.

matt3141 Over a year ago

Sorry guys, I'm always surprised when I rediscover that Python lacks the Ternary Operator.

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